# Factoring a third degree polynomial

Ibraheem

## Homework Statement

Factor out the polynomial and find its solutions x^3-5x^2+7x-12[/B]

## The Attempt at a Solution

I tried to factor it, but I'm stuck in this step x^2(x-5)+7(x-5)+23= 0. I graphed the equation, and I know there is two imaginary solutions and one real positive integer which means there's a linear factor that I can't reach to without graphing the polynomial . So how do I factor the polynomial without referring to the graph? [/B]

Homework Helper
Try to find the real root among the dividers of the constant term. (what about 4?) It also helps if you find the extrema of the function.

. So how do I factor the polynomial without referring to the graph?

You could observe the polynomial is negative at x = 0 and (using the way you rewrote it) that it is positive at x = 5. So without graphing it, you can tell the graph crosses the x-axis between x = 0 and x = 5. Look for an integer root that is between 0 and 5 by trial and error.

Ibraheem
Staff Emeritus
Homework Helper

## Homework Statement

Factor out the polynomial and find its solutions x^3-5x^2+7x-12[/B]

## The Attempt at a Solution

I tried to factor it, but I'm stuck in this step x^2(x-5)+7(x-5)+23= 0. I graphed the equation, and I know there is two imaginary solutions and one real positive integer which means there's a linear factor that I can't reach to without graphing the polynomial . So how do I factor the polynomial without referring to the graph? [/B]

Typically, one uses Descartes rule of signs to determine the number of real roots and the rational root theorem to deduce trial solutions for polynomials.

http://en.wikipedia.org/wiki/Descartes'_rule_of_signs

http://en.wikipedia.org/wiki/Rational_root_theorem

Ibraheem
Homework Helper
Dearly Missed

## Homework Statement

Factor out the polynomial and find its solutions x^3-5x^2+7x-12[/B]

## The Attempt at a Solution

I tried to factor it, but I'm stuck in this step x^2(x-5)+7(x-5)+23= 0. I graphed the equation, and I know there is two imaginary solutions and one real positive integer which means there's a linear factor that I can't reach to without graphing the polynomial . So how do I factor the polynomial without referring to the graph? [/B]

Use the Rational Root Theorem: see, eg.,
http://en.wikipedia.org/wiki/Rational_root_theorem
or
http://www.purplemath.com/modules/rtnlroot.htm

Note added in edit: I see that the rational root theorem has already been suggested to you.

Last edited:
Ibraheem
Homework Helper
Gold Member
These problems given with integer coefficients usually have integer solutions (though that's a fact about the problems they give you, not a fact about polynomials with integer coefficients in general). So look at the last term, ±12 or ± one or more of its factors you expect to be a solution. Try. Then there's a bit more to it but which is, or will become, fairly obvious.